Find $\lim_{x\to 0}\dfrac{2e^{2x}-2}{x^{2}+x}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac14$ (Choice B) B $1$ (Choice C) C $4$ (Choice D) D The limit doesn't exist.
Answer: Substituting $x=0$ into $\dfrac{2e^{2x}-2}{x^{2}+x}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 0}\dfrac{2e^{2x}-2}{x^{2}+x} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[2e^{2x}-2]}{\dfrac{d}{dx}[x^{2}+x]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{4e^{2x}}{2x+1} \\\\ &=\dfrac{4e^{2(0)}}{2(0)+1} \gray{\text{Substitution}} \\\\ &=4 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[2e^{2x}-2]}{\dfrac{d}{dx}[x^{2}+x]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{2e^{2x}-2}{x^{2}+x}=4$.